Differentiate between the principal of estimation of nitrogen in an organic compound by (i) Kjeldehl's method and (ii) Dumas method.

Explanation

Kjeldahl.s method : This is another method to estimate nitrogen. In this method, the compound is heated with concentrated sulphuric acid in the presence of small amount of `Cu SO _(4).` Nitrogen is quantitatively converted into ammonium sulphate. The contents of the flask are transferred to another flask and heated with excess of sodium hydroxide solution to liberate ammonia gas. Ammonia gas so liberated is passed and absorbed in a known volume of known concentrated sulphuric acid that is relatively more in amount than that is required to neutralise `NH_(3)` gas, Now, the excess of acid remained after the neutralisation by `NH_(3)` is titrated against a standard solution of alkali. From the above, the amount of `H_(2)SO_(4)` used to neutralise `NH_(3)` is calculated.. From this, the mass of ammonia formed is calculated and from that percentage of nitrogen is calculated.
Organic compound `+ H _(2) SO _(4) to (NH) _(4) SO _(4)`
`(NH _(4)) _(2) SO _(4) + 2 NaOH to N a _(2) SO _(4) + 2 H _(2) O + 2 NH _(3)`
`2 NH _(3) + H_(2) SO_(4) to (NH_(4)) _(2) SO _(4)`
Calculation:
Let hte mass of organic compoind taken be .a. g.
Let the volume of `H _(2) SO _(4)` initially taken be `.V _(ml).` and its molarity M.
After passing the `NH_(3)` gas into the above acid, if the remaining acid is titrated with molar NaoH and it consumes `V _(1)` ml. of NaOH for complete neutrallsation, then from the formula
`(M V _(1))/( n _(1) ) ( Na OH) = ( MV _(2))/( n _(2)) ( H _(2) SO _(4))`
From the stoichiometric equation
`2 NaOH + H _(2) SO _(4) to Na _(2) SO _(4) + 2 H _(2) O`
`n _(1) =` number of moles of `Na OH = 2 , n _(2)=` number of moles of `H _(2) SO_(4) =1`
`( M V _(1))/(2) = ( MV _(2))/( l ) or V _(2) = ( V _(l))/(2) ml`
Therefore, the volume of `H _(2) SO _(4)` neutralised by `NH _(2)is [ V - ( V _(1))/(2) ] ml.`
(or) it is equal to `2 [ V - ( V _(1))/(2)]` ml. M molar `NH_(3)]` solution.
1000 ml. of `1M NH_(3)` solution contains 17g. of `NH_(3) or 14g. of N _(2)`
`2 [ V - ( V _(1))/( 2 ) ml`of .M. ` NH _(3)` solutions contains .......?
` (1. 4 xx M xx 2 [ V - (V _(1))/(2) ])/( 1000) g. ` of nitrogen
Percentage of nitrogen `= (14 xx M xx 2 [ V - (V _(1))/( 2) ])/( 1000 ) xx (100)/(a) `